Bonjour
Yannis27
Formules utilisées :
[tex](a+b)^3=a^3+3a^2b+3ab^2+b^3\\\\a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]
[tex]1)\ (x+y)^3+z^3=x^3+3x^2y+3xy^2+y^3+z^3\\\\x^3+y^3+z^3=(x+y)^3+z^3-3x^2y-3xy^2\\\\x^3+y^3+z^3=[(x+y)+z][(x+y)^2-(x+y)z+z^2]-3x^2y-3xy^2\\\\x^3+y^3+z^3=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3x^2y-3xy^2\\\\x^3+y^3+z^3=0\times[(x+y)^2-(x+y)z+z^2]-3x^2y-3xy^2\\\\x^3+y^3+z^3=-3x^2y-3xy^2\\\\x^3+y^3+z^3=-3xy(x+y)\\\\x^3+y^3+z^3=-3xy\times(-z)\ car\ x+y+y=0\Longrightarrow x+y=-z\\\\\boxed{x^3+y^3+z^3=3xyz}[/tex]
[tex]2)\ x\in[\dfrac{1}{2}\ ;\ \dfrac{3}{2}]\Longleftrightarrow\dfrac{1}{2}\le x\le\dfrac{3}{2}\Longrightarrow(\dfrac{1}{2})^2\le x^2\le(\dfrac{3}{2})^2[/tex]
car la fonction "carré" est croissante sur [0;+oo[.
D'où
[tex]\boxed{\dfrac{1}{4}\le f(x)\le\dfrac{9}{4}}\\\\\boxed{f(x)\in[\dfrac{1}{4}\ ;\ \dfrac{9}{4}]}[/tex]
et
[tex](\dfrac{1}{4}\le x^2\le\dfrac{9}{4}\ \ et\ \ \dfrac{1}{2}\le x\le\dfrac{3}{2})\\\\\Longrightarrow\dfrac{1}{4}+\dfrac{1}{2}+3\le x^2+x+3\le\dfrac{9}{4}+\dfrac{3}{2}+3\\\\\Longrightarrow\boxed{\dfrac{15}{4}\le g(x)\le\dfrac{27}{4}}\\\\\Longrightarrow\boxed{g(x)\in[\dfrac{15}{4}\ ;\ \dfrac{27}{4}]}[/tex]
3) Résoudre dans R :
|2x - 1| = |3x + 1|
2x - 1 = 3x + 1 ou 2x - 1 = -3x - 1
2x - 3x = 1 + 1 ou 2x + 3x = -1 + 1
-x = 2 ou 5x = 0
x = -2 ou x = 0
L'ensemble des solutions de l'équation est S = { -2 ; 0}
